Question

Q. Solve the following differential equation:
(D3−D2+3D+5)y=exsin2x

Answer 1

Answer on Question #64621 – Math – Differential Equations

Question

Solve the following differential equation:

$$(D3 - D2 + 3D + 5)y = \text{exsin}2x$$

Solution

To solve the inhomogeneous linear differential equation (DE)

$$y''' - y'' + 3y' + 5y = e^x \sin 2x$$

we must

1) find the complementary function $y_c$ that is the general solution of the associated homogeneous DE

$$y''' - y'' + 3y' + 5y = 0;$$

2) find any particular solution $y_p$ of the nonhomogeneous equation

$$y''' - y'' + 3y' + 5y = e^x \sin 2x.$$

The general solution of the equation (1) is $y = y_c + y_p$.

1) To solve the associated homogeneous DE

$$y''' - y'' + 3y' + 5y = 0$$

we substitute into equation the solution as the exponential function $y = e^{mx}$. We get

$$m^3 e^{mx} - m^2 e^{mx} + 3m e^{mx} + 5e^{mx} = 0$$

or

$$(m^3 - m^2 + 3m + 5) e^{mx} = 0.$$

Since $e^{mx}$ is not zero for real values of $x$, the last equation has only the roots which are the solutions of the third-degree polynomial equation:

$$m^3 - m^2 + 3m + 5 = 0.$$

This equation is called the auxiliary equation of the differential equation (2).

It is obvious that one of the roots is $m = -1$. This can be verified by substituting $m = -1$ in equation

$$(-1)^3 - (-1)^2 + 3(-1) + 5 = -1 - 1 - 3 + 5 = 0.$$

To find other roots divide $m^3 - m^2 + 3m + 5$ by $m + 1$

$$\begin{array}{c} \dfrac{m^2 - 2m + 5}{m + 1 \mid m^3 - m^2 + 3m + 5} \\ \dfrac{m^3 + m^2}{-2m^2 + 3m} \\ \dfrac{-2m^2 - 2m}{5m + 5} \\ \dfrac{5m + 5}{0} \end{array}$$ $$y_p = x e^x (A \cos(2x) + B \sin(2x)),$$ $$y_p' = A e^x \cos 2x + A x e^x \cos 2x - 2A x e^x \sin 2x + B e^x \sin 2x + B x e^x \sin 2x + 2B x e^x \cos 2x,$$ y_p''_p = A e^x \cos 2x - 2A e^x \sin 2x + A e^x \cos 2x + A x e^x \cos 2x - 2A x e^x \sin 2x - 2A e^x \sin 2x - 2A x e^x \sin 2x $$\begin{array}{l} -4Axe^x \cos 2x + Be^x \sin 2x + 2Be^x \cos 2x + Be^x \sin 2x + Bxe^x \sin 2x + 2Bxe^x \cos 2x \\ + 2Be^x \cos 2x + \\ + 2Bxe^x \cos 2x - 4Bxe^x \sin 2x, \end{array}$$

or

$$\begin{array}{l} y''_p = 2Ae^x \cos 2x - 4Ae^x \sin 2x - 3Axe^x \cos 2x - 4Axe^x \sin 2x + \\ + 2Be^x \sin 2x + 4Be^x \cos 2x - 3Bxe^x \sin 2x + 4Bxe^x \cos 2x \end{array}$$ $$\begin{array}{l} y''_p = 2Ae^x \cos 2x - 4Ae^x \sin 2x - 4Ae^x \sin 2x - 8Ae^x \cos 2x - 3Ae^x \cos 2x - 3Axe^x \cos 2x \\ + 6Axe^x \sin 2x - 4Ae^x \sin 2x - 4Ae^x \sin 2x - 8Axe^x \cos 2x + 2Be^x \sin 2x + 4Be^x \cos 2x \\ + 4Be^x \cos 2x - \\ -8Be^x \sin 2x - 3Be^x \sin 2x - 3Bxe^x \sin 2x - 6Bxe^x \cos 2x + 4Be^x \cos 2x + 4Bxe^x \cos 2x \\ - 8Bxe^x \sin 2x \end{array}$$

or

$$\begin{array}{l} y''_p = -9Ae^x \cos 2x - 12Ae^x \sin 2x - 11Axe^x \cos 2x + 2Axe^x \sin 2x - \\ -9Be^x \sin 2x + 12Be^x \cos 2x - 11Bxe^x \sin 2x - 2Bxe^x \cos 2x \end{array}$$

substitute the derivatives into the equation (1)

$$\begin{array}{l} y''' - y'' + 3y' + 5y = -9Ae^x \cos 2x - 12Ae^x \sin 2x - 11Axe^x \cos 2x + 2Axe^x \sin 2x - \\ -9Be^x \sin 2x + 12Be^x \cos 2x - 11Bxe^x \sin 2x - 2Bxe^x \cos 2x - (2Ae^x \cos 2x - 4Ae^x \sin 2x - \\ -3Axe^x \cos 2x - 4Axe^x \sin 2x + 2Be^x \sin 2x + 4Be^x \cos 2x - 3Bxe^x \sin 2x + 4Bxe^x \cos 2x) + \\ + 3(Ae^x \cos 2x + Axe^x \cos 2x - 2Axe^x \sin 2x + Be^x \sin 2x + Bxe^x \sin 2x + 2Bxe^x \cos 2x) + \\ + 5(Axe^x \cos 2x + Bxe^x \sin 2x) = e^x \sin 2x \end{array}$$

Crossed out vanished terms and we have

$$\begin{array}{l} -9Ae^x \cos 2x - 12Ae^x \sin 2x - 9Be^x \sin 2x + 12Be^x \cos 2x - (2Ae^x \cos 2x - 4Ae^x \sin 2x - \\ + 2Be^x \sin 2x + 4Be^x \cos 2x) + 3(Ae^x \cos 2x + Be^x \sin 2x) = e^x \sin 2x \end{array}$$

Equating the coefficients of corresponding terms, we find

$$\begin{array}{l} \left\{ \begin{array}{l} -9A + 12B - 2A - 4B + 3A = -6A + 6B = 0 \\ -12A - 9B + 4A - 2B + 3B = -8A - 8B = 1 \end{array} \right. \end{array}$$

Then we have $A = B = -\frac{1}{16}$ and finally we get the general solution of differential equation (1)

$$y = c_1 e^{-x} + c_2 e^x \cos 2x + c_3 e^x \sin 2x - \frac{1}{16} x e^x (\cos 2x + \sin 2x)$$

Answer: $y = c_1 e^{-x} + c_2 e^x \cos 2x + c_3 e^x \sin 2x - \frac{1}{16} x e^x (\cos 2x + \sin 2x)$.

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